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trig function?

can somebody just provide me with how to figure this out? i don't need the answers an explanation will be great. Thank you

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4 Answers

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  • TomV
    Lv 7
    2 months ago
    Favorite Answer

    Use the relationships:

    sinΘ = y/r 

    cosΘ = x/r

    tanΘ = sinΘ/cosΘ = y/x

    secΘ = 1/cosΘ = r/x

    cscΘ = 1/sinΘ = r/y

    cotΘ = 1/tanΘ = x/y

    and the fact that in Quadrant 2, y is positive and x is negative.

    Given: sinΘ = 2/7 from which we know that r = 7 and y = 2

    Calculate x = ±√(r² - y²) [choose the negative value because of Q2]

    x = -√(49-4) = -√45 = -3√5

    Use those values to calculate the remaining functions.

    cosΘ = -3√5/7

    tanΘ = 2/(-3√5) = -2√5/15

    secΘ = 7/(-3√5) = -7√5/15

    cscΘ = 7/2

    tanΘ = -3√5/2

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  • 2 months ago

    By Pythagoras we need to find the base(Adjacent side). 

    Hence 

    7^2 = 2^2 + b^2

    b^2 = 7^2 - 2^2 

    b^2 = (7 - 2)(7 + 2) 

    b^2 = 5 * 9 = 45 

    b = 6.7082... 

    b = -6.7082 .. is negative because it is in the 2nd quadrant. 

    Hence

    Cos(Th) = 6.7082.../7 

    Tan(Th) = 2/6.7082...

    Sec(th) = 1/Cos(Th) = 7/6.7082...

    Csc(Th) = 1/Sin(Th) = 7/2 

    Cot(Th) = 1/Tan(Th) = 6.7082../2

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  • Pope
    Lv 7
    2 months ago

    cos²(θ) = 1 - sin²(θ)

    cos²(θ) = 1 - (2/7)²

    cos²(θ) = 45/49

    cos(θ) = -5√(3)/7 ... negative because θ is in quadrant II

    tan(θ) = sin(θ)/cos(θ)

    sec(θ) = 1/cos(θ)

    csc(θ) = 1/sin(θ)

    cot(θ) = cos(θ)/sin(θ)

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  • Ash
    Lv 7
    2 months ago

    (a) Remember the identity involving sin and cos....

    sin²θ+cos²θ = 1

    cos²θ = 1- sin²θ

    Now plug in sinθ and find cosθ

    b) tanθ = sinθ / cosθ

    c) secθ = 1/cosθ

    d) cscθ = 1/sinθ

    e) cotθ = 1/tanθ

    Good luck !

    • olayda2 months agoReport

      thank you!

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